$f(x) = \begin{cases} -3\sqrt{x } & \text{for} ~~~~x\gt{1} \\ x-3& \text{for} ~~~~ x \leq1\end{cases}$ Evaluate the definite integral. $\int^4_{-4}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $-16$ (Choice B) B $-24$ (Choice C) C $ -\dfrac{73}{2}$ (Choice D) D $35$
Explanation: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^4_{-4}f(x)\,dx$ $= \int^4_{1}f(x)\,dx + \int^{1}_{-4}f(x)\,dx~~~~~~$ [Why did we split at 1?] $= \int^4_{1}-3\sqrt{x}\,dx + \int^1_{-4}(x-3)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^4_{1}-3\sqrt{x}\,dx &=-2x^\frac32\Bigg|^4_{{1}} \\\\ &= \left[-2 ( 4)^\frac32 \right] - \left[-2 ({1})^\frac32\right] \\\\ &= \left[-16\right] -\left[-2 \right] \\\\ &= {-14}\end{aligned}$ The second definite integral: $\begin{aligned} \int^1_{-4}(x-3)\,dx &=\left(\dfrac12x^2-3x\right)\Bigg|^1_{{-4}} \\\\ &= \left[\dfrac12 \cdot( 1)^2 -3\cdot(1) \right] - \left[\dfrac12 \cdot( {-4})^2 -3\cdot({-4})\right] \\\\ &= \left[-\dfrac{5}{2}\right] -\left[20 \right] \\\\ &= {-\dfrac{45}2}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^4_{1}-3\sqrt{x}\,dx + \int^1_{-4}(x-3)\,dx$ $ ={-14} + \left({-\dfrac{45}{2}}\right)$ $ = -\dfrac{73}{2}$ The answer $\int^4_{-2}f(x)\,dx = -\dfrac{73}{2}$